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In a refrigerator refrigerant-134a enters the compressor as superheated vapor at a rate of 0.12 kg/s with h = 246.36 kJ/kg and it leaves with h = 288.53 kJ/kg. The isentropic enthalpy of refrigerant leaving compressor is hs = 281.16 kJ/kg. The refrigerant is cooled in the condenser and it leaves the condenser in subcooled region with h = 84.98 kJ/kg. It is then throttled to 0.15 MPa. The isentropic efficiency of compressor and COP of refrigerator respectively are
  • a)
    50.2%, 4.52
  • b)
    59.3%, 2.23
  • c)
    82.5%, 3.83
  • d)
    72.6%, 5.40
Correct answer is option 'C'. Can you explain this answer?
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In a refrigerator refrigerant-134aenters the compressor assuperheated ...
 
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In a refrigerator refrigerant-134aenters the compressor assuperheated ...
Given data:
- Mass flow rate of refrigerant-134a, m = 0.12 kg/s
- Enthalpy at the compressor inlet, h1 = 246.36 kJ/kg
- Enthalpy at the compressor outlet, h2 = 288.53 kJ/kg
- Isentropic enthalpy at the compressor outlet, hs2 = 281.16 kJ/kg
- Enthalpy at the condenser outlet, h3 = 84.98 kJ/kg
- Pressure after throttling, P4 = 0.15 MPa

Calculating the isentropic efficiency of the compressor:
The isentropic efficiency of the compressor can be calculated using the following formula:
η = (hs2 - h1) / (h2 - h1) * 100%

Substituting the given values:
η = (281.16 - 246.36) / (288.53 - 246.36) * 100%
η ≈ 82.5%

Calculating the Coefficient of Performance (COP) of the refrigerator:
The COP of the refrigerator can be calculated using the following formula:
COP = Qc / W

Where Qc is the heat removed in the condenser and W is the work done by the compressor.

Calculating the heat removed in the condenser:
The heat removed in the condenser can be calculated using the following formula:
Qc = m * (h2 - h3)

Substituting the given values:
Qc = 0.12 * (288.53 - 84.98)
Qc ≈ 22.2 kW

Calculating the work done by the compressor:
The work done by the compressor can be calculated using the following formula:
W = m * (h2 - h1)

Substituting the given values:
W = 0.12 * (288.53 - 246.36)
W ≈ 5.06 kW

Calculating the COP:
COP = Qc / W
COP = 22.2 / 5.06
COP ≈ 4.38

Therefore, the isentropic efficiency of the compressor is approximately 82.5% and the COP of the refrigerator is approximately 4.38. The closest option to these values is option 'C' with an isentropic efficiency of 82.5% and a COP of 3.83.
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In a refrigerator refrigerant-134aenters the compressor assuperheated vapor at a rate of 0.12kg/s with h = 246.36 kJ/kg and itleaves with h = 288.53 kJ/kg. Theisentropic enthalpy of refrigerantleaving compressor is hs = 281.16kJ/kg. The refrigerant is cooled inthe condenser and it leaves thecondenser in subcooled region withh = 84.98 kJ/kg. It is then throttled to 0.15 MPa. The isentropicefficiency of compressor and COPof refrigerator respectively area)50.2%, 4.52b)59.3%, 2.23c)82.5%, 3.83d)72.6%, 5.40Correct answer is option 'C'. Can you explain this answer?
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In a refrigerator refrigerant-134aenters the compressor assuperheated vapor at a rate of 0.12kg/s with h = 246.36 kJ/kg and itleaves with h = 288.53 kJ/kg. Theisentropic enthalpy of refrigerantleaving compressor is hs = 281.16kJ/kg. The refrigerant is cooled inthe condenser and it leaves thecondenser in subcooled region withh = 84.98 kJ/kg. It is then throttled to 0.15 MPa. The isentropicefficiency of compressor and COPof refrigerator respectively area)50.2%, 4.52b)59.3%, 2.23c)82.5%, 3.83d)72.6%, 5.40Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In a refrigerator refrigerant-134aenters the compressor assuperheated vapor at a rate of 0.12kg/s with h = 246.36 kJ/kg and itleaves with h = 288.53 kJ/kg. Theisentropic enthalpy of refrigerantleaving compressor is hs = 281.16kJ/kg. The refrigerant is cooled inthe condenser and it leaves thecondenser in subcooled region withh = 84.98 kJ/kg. It is then throttled to 0.15 MPa. The isentropicefficiency of compressor and COPof refrigerator respectively area)50.2%, 4.52b)59.3%, 2.23c)82.5%, 3.83d)72.6%, 5.40Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a refrigerator refrigerant-134aenters the compressor assuperheated vapor at a rate of 0.12kg/s with h = 246.36 kJ/kg and itleaves with h = 288.53 kJ/kg. Theisentropic enthalpy of refrigerantleaving compressor is hs = 281.16kJ/kg. The refrigerant is cooled inthe condenser and it leaves thecondenser in subcooled region withh = 84.98 kJ/kg. It is then throttled to 0.15 MPa. The isentropicefficiency of compressor and COPof refrigerator respectively area)50.2%, 4.52b)59.3%, 2.23c)82.5%, 3.83d)72.6%, 5.40Correct answer is option 'C'. Can you explain this answer?.
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